 # How can I prove <msubsup> <mi mathvariant="normal">&#x03A3;<!-- Σ --> <mrow class="MJX Villaretq0 2022-06-27 Answered
How can I prove ${\mathrm{\Sigma }}_{i=1}^{n-1}i={\left(}_{2}^{n}\right)$
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Step 1
That $\sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2}$ can be proved by induction.
The case $k=1$ is trivial and assume it also holds for $k=n-1$. But then $\sum _{i=1}^{n}i=\frac{\left(n-1\right)n}{2}+n=\frac{n\left(n-1\right)+2n}{2}=\frac{n\left(n+1\right)}{2}$. Hence $\sum _{i=1}^{n-1}i=\frac{n\left(n-1\right)}{2}$.
Step 2
But by definition $\left(\genfrac{}{}{0}{}{n}{2}\right)=\frac{n!}{2!\left(n-2\right)!}=\frac{n\left(n-1\right)\left(n-2\right)!}{2\left(n-2\right)!}=\frac{n\left(n-1\right)}{2}$
Therefore we can conclude that $\sum _{i=1}^{n-1}i=\left(\genfrac{}{}{0}{}{n}{2}\right)$

We have step-by-step solutions for your answer! Jaqueline Kirby
Explanation:
Here’s a combinatorial proof: To choose two out of n elements (in $\left(\genfrac{}{}{0}{}{n}{2}\right)$ ways), first choose the $\left(i+1\right)$-th element and then choose one of the i elements preceding it.

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